Question

In 2015, the average cost of attendance at public 2-year colleges and technical schools across the United States was $12,260. A researcher wants to know if the 2-year colleges in Michigan are less expensive.

1. A QQ-plot of the cost of attendance at public 2-year colleges and technical schools is shown below. Based on this plot, is it reasonable to use this data to evaluate the research question? Why or why not?

? Yes No  because  ? The population data distribution seems approximately normal. The population data distribution does not seem approximately normal. A sample size of 15 is too small. A sample size of 15 is very large.   

Normal Q-Q Plot 16000 14000 Sample Quantiles 12000 10000 8000 ooooo E Theoretical Quantiles

The researcher decides to conduct a hypothesis test. They take a random sample of 15 such schools in Michigan and find a sample average of 10072 with a standard deviation of 1464. Conduct a hypothesis test to determine if the average cost of attendance at 2-year colleges in Michigan is lower than the national average. Round all numeric results to 4 decimal places.

2. Which set of hypotheses should the researcher use?
A. ?0H0: ?μ = 12,260 vs.  ??HA: ?μ > 12,260
B. ?0H0: ?μ = 12,260 vs.  ??HA: ?μ < 12,260
C. ?0H0: ?μ = 12,260 vs.  ??HA: ?≠μ≠ 12,260

3. Calculate the test statistic.

4. Calculate the p-value.

5. Based on the p-value, we have:
A. extremely strong evidence
B. strong evidence
C. very strong evidence
D. some evidence
E. little evidence
that the null model is not a good fit for our observed data.

6. Calculate a 95% confidence interval for the mean cost of attendance at 2-year colleges in Michigan. ($  , $ )

Answer 1

1) As can be seen from the QQ plot, majority of the points lie on the straight line.

Thus, the data is from a normally distributed population.

Hence, it is reasonable to use this data to evaluate the research question.

Answer

Yes

because The population data distribution seems approximately normal.

The research question is to test if the 2-year colleges in Michigan are less expensive that is their expense is less than the population average expense of $12,260

Data Summary Standard Sample Size Deviation (u) (X) 15 (n) 1464 (s) Mean 12260 10072 Population Sample

2)

Since we have to test if the Michigan colleges are less expensive, the null and alternative hypotheses are

Ho : μ = 12260       μ is the population mean cost of attendance of 2 year colleges across the USA
HA : μ < 12260      

Answer

B.    ?0: ? = 12,260 vs.  ??: ? < 12,260    

3)   test statistic

We use T test since population standard deviation is unknown, and sample size is small              
Using the below formula, we get the test statistic              
              
    t = -5.7883     
test statistic t = -5.7883         

K-H s 10072 - 12260 1464 15

4) p-value

Degrees of Freedom              
df = n - 1 = 14         

Since the alternative hypothesis is "less than", this is a left tailed test

For t = -5.7883 df = 14 we find the Left Tailed p-value using Excel function t.dist              

p-value = t.dist(-5.7883, 14, TRUE)              
p-value = 0    (p-value is almost 0)             
              

5)    We test for 1% level of significance

that is α = 0.01
Decision              
0 < 0.01             
that is p-value < α              
Hence we REJECT Ho      

Since p-value is zero and less than 0.01 the evidence is very strong     

Answer
Based on the p-value, we have:

C. very strong evidence

that the null model is not a good fit for our observed data.

6) 95% confidence interval

For 95% Confidence interval
α = 0.05,      α/2 = 0.025
From t tables of Excel function T.INV.2T (α, degrees of freedom) we find the t value
t = T.INV.2T(0.05, 14) = 2.145
We take the positive value of t

Confidence interval is given by

Itx 10072 + 2.145 x 146 V15

= (9261.1832, 10882.8168)

= ($9261.18, $10882.82)          ...Rounding to 2 decimals

95% Confidence interval for the mean cost of attendance at 2-year colleges in Michigan ($9261.18, $10882.82)